CPhill

What  all  4 took  must be all the berries in the  bowl....call this N

Billy took (1/2)N + 2

Bob  took  (1/3)  [ (1/2)N  + 2 ] + 2

Joe took   (1/3)   [  (1/3)  [ (1/2)N  + 2 ] + 2 ]  -1

Mom  took 1

So we  have  this rather convoluted equation

[ (1/2) N + 2  ]+    (1/3)  [ (1/2)N  + 2 ] + 2  +  (1/3)   [  (1/3)  [ (1/2)N  + 2 ] + 2 ]  -1 + 1  =  N

Because  this is a little tedious  ( but not impossible) to solve, I'll get a little  help from WolframAlpha

N  = 20

Billy took  (1/2) (20) + 2  = 10 + 2  =   12

Bob took  (1/3) of what Billy took + 2 = (1/3)12+ 2  =  4 + 2  =  6

Joe took   (1/3)  of what Bob took and put one  back  =(1/3)(6)  - 1   =   2  - 1  =  1

And Mom took 1

And   12  + 6 + 1 + 1    = 20 berries in the  bowl originally

It  will pass through 13 lattice points

To  see this, it will  pass through a lattice point   at  every x value from  0  to  12

And  these will  be paired with y values  36, 33, 30 , 27.......0

So....13 lattice points

4x + 6y + 13 + 2x   -2y   simplifies to

6x + 4y + 13

So

x^2 + y^2  ≤  6x + 4y  + 13

x^2  -6x  + y^2 -4y  ≤ 13       complete the square on x and  y

x^2  - 6x + 9  +  y^2  - 4y + 4 ≤ 13 + 9 + 4

(x - 3)^2  + (y - 2)^2  ≤  26

This is just the  area  inside  the circle  with a center of  ( 3, 2)   and a radius of  sqrt (26)

See here  : https://www.desmos.com/calculator/0c8kf8nfll

The area  is    pi (sqrt (26))^2 =    26 pi

True....equal...

Pure  acid   =100%  acid  =  1

So...let  x  be the amout of pure acid to  be added and we  have that

1x +.10 *( 90 - x)  =  ( .58)(90)

x  + 9  - .10x  =  52.2

.90x  =  52.2  - 9

.90x  =  43.2      divide both sides by  .90

x  = 43.2  / .90  =  48   L

That is pretty cool   !!!!

25/5  =   5 zeroes

25/5^2 =  25/25  = 1  more zero

So

5 +  1    =  6 zeroes

No odd ones are  divisible  by  8

So   we  have

8^2  = 64   = 8 * 8

12^2  = 144  = 8 * 18

16^2   = 256  = 8 * 32

20^2  = 400  =  8 * 50

24^2  = 576  =  8 * 72

28^2    =  784  =  8 * 96

b = 50 (2)^h

 If the bacteria continue to grow at the same rate, there will be 800 bacteria after 4 hours. 

Need to know how many  cups  the full pitcher holds, Alston, to give a definitive answer

However....let's suppose that  the full pitcher holds  N cups

Ater 1/8N  is  poured out  .....there remains   (7/8)N  cups

The  amt poured out to water the plant

So  (1/4) (7/8)N   =  7/32N    

So....the amt  that remains   =   (7/8  - 7/32)N  =  ( 28/32  - 7/32)N  =  (21/32)N